Subtract the following rational expressions. $\dfrac{-5x}{8x+7}-\dfrac{6x^3}{3x+1}=$
Solution: We can subtract two rational expressions whose denominators are equal by subtracting the numerators and keeping the denominator the same. [Does this fit with how we subtract rational numbers?] When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator. Since the two denominators do not share any common factors, the common denominator is simply the product of these two denominators: $({8x+7})\cdot({3x+1})$. Let's manipulate the expressions to have that denominator: $\begin{aligned} &\phantom{=}\dfrac{-5x}{{8x+7}}-\dfrac{6x^3}{{3x+1}} \\\\ &=\dfrac{-5x\cdot({3x+1})}{({8x+7})\cdot({3x+1})}-\dfrac{6x^3\cdot({8x+7})}{({3x+1})\cdot({8x+7})} \end{aligned}$ [Why did we do that?] Now that both denominators are the same, let's subtract! $\begin{aligned} &\phantom{=}\dfrac{-5x\cdot(3x+1)}{(8x+7)\cdot(3x+1)}-\dfrac{6x^3\cdot(8x+7)}{(3x+1!\cdot(8x+7)} \\\\ &=\dfrac{-5x\cdot(3x+1)-6x^3\cdot(8x+7)}{(8x+7)(3x+1)} \\\\ &=\dfrac{-15x^2-5x-48x^4-42x^3}{(8x+7)(3x+1)} \\\\ &=\dfrac{-48x^4-42x^3-15x^2-5x}{(8x+7)(3x+1)} \end{aligned}$ In conclusion, $\dfrac{-5x}{8x+7}-\dfrac{6x^3}{3x+1}=\dfrac{-48x^4-42x^3-15x^2-5x}{(8x+7)(3x+1)}$